Multivariable continuity and partial derivatives question P1 : Given $f(x) = x$ if $y=0$,  $x - (y^3)\sin(1/y)$ when $y$ is not equal to $0$. Is $f$ continous and differentiable at at $(0,0)$?
P2. Given $f = x  \sqrt{x^2 + y^2} /  |xy|$  when $x$ is not equal to $0$ 
and $0$ when $x=0$. Then the value of partial $f$ w.r.t. $x$ and partial $f$ w.r.t. $y$ at $(0,0)$ are?
 A: For your first question, notice that $f_x(x,y)=1$ (for all $y$). This is continuous everywhere. 
$f_y(x,y)$ is definitely continuous for $y \not=0$ (just differentiate your formula for $f$ when $y \not=0$):
$$f_y(x,y) = -3y^2\sin(1/y) -y^3\cos(1/y)(-1/y^2) = -3y^2\sin(1/y)+y\cos(1/y)$$
The case "$f_y$ when $y=0$" requires more care:
$$f_y(x,0) = \lim\limits_{h \to 0} \dfrac{f(x,0+h)-f(x,0)}{h} = \lim\limits_{h \to 0} \dfrac{x-h^3\sin(1/h) - x}{h} = \lim\limits_{h \to 0} h^2\sin(1/h) = 0.$$
Finally, notice that $\lim\limits_{y \to 0} f_y(x,y) = 0 = f_y(x,0)$. 
In the end, we see that $f_x$ and $f_y$ exist and are continuous everywhere. Continuity of first partials of $f$ then implies differentiability of $f$ which implies continuity of $f$.
Your second function has issues. It isn't defined near $(0,0)$. Why? Notice that $f(x,y)=x\sqrt{x^2+y^2}/|xy|$ isn't defined when $x=0$ (which you've handled) but it's also undefined when $y=0$.
If you change the definition of (the second) $f(x,y)$ so that $f(x,y)=0$ for $x=0$ or $y=0$, then the first partials at $(0,0)$ are 0 (use the limit definition).
Edit: Details about the second computation...
$$f_x(0,0) = \lim\limits_{h\to 0} \dfrac{f(0+h,0)-f(0,0)}{h} = \lim\limits_{h \to 0} \dfrac{f(h,0)-f(0,0)}{h} = \lim\limits_{h \to 0} \dfrac{0-0}{h} = 0$$
since $f(\mbox{anything},0)=0$ according to my "fixed" definition of $f(x,y)$.
